Why not to use Bayes theorem in the form $p(\theta | x) = \frac{L(\theta | x) p(\theta)}{p(x)}$? There are a lot of questions (like this) about some ambiguity with Bayesian formula in continuous case.
$$p(\theta | x) = \frac{p(x | \theta) \cdot p(\theta)}{p(x)}$$
Oftentimes, confusion arises from the fact that definition of conditional distribution $f(variable | parameter) $ is explained as $f$ being function of $variable$ given fixed $parameter$.
Alongside with that, there is an equivalence principle stating that likelihood can be written as:
$$ L(\theta | x) = p(x | \theta)$$
So why not to use Bayes rule for distributions in the following form:
$$p(\theta | x) = \frac{L(\theta | x) \cdot p(\theta)}{p(x)}$$
to emphasize that we are dealing with functions of $\theta$ given observed data $x$, and that the respective term is likelihood (at least, starting with $L$)?
Is this a matter of tradition, or is there something more fundamental in this practice?
 A: There are two basic results from probability that are at work in Bayes' theorem. One is a way of rewriting a joint probability density function:
$$p(x,\,y)=p(x\,|\,y)p(y).$$
The other is a formula for computing a conditional probability density function:
$$p(y\,|\,x)=\frac{p(x,\,y)}{p(x)}.$$
Bayes' theorem just stitches these two things together:
$$p(\theta\,|\,x)=\frac{p(x,\,\theta)}{p(x)}=\frac{p(x\,|\,\theta)p(\theta)}{p(x)}$$
So both the data $x$ and the parameters $\theta$ are random variables with joint pdf
$$p(x,\,\theta)=p(x\,|\,\theta)p(\theta),$$
and that's what shows up in the numerator in Bayes' theorem. So writing the likelihood as a conditional probability density instead of as a function $L$ of the parameters makes clear the basic probability at play.
That all said, you'll see people use either, like here or here.
A: The likelihood function is merely proportional to the sampling density, in the sense that you have $L_x(\theta) = k(x) \cdot p(x|\theta)$ for some constant $k(x) > 0$ (though you should note that the likelihood is a function of the parameter, not the data).  If you want to use this in your expression for Bayes theorem then you need to include the same scaling constant in the denominator:
$$p(\theta|x) = \frac{L_x(\theta) \cdot p(\theta)}{k(x) \cdot p(x)} = \frac{L_x(\theta) \cdot p(\theta)}{\int L_x(\theta) \cdot p(\theta) \ d \theta} \propto L_x(\theta) \cdot p(\theta).$$
If you instead use the formula you have proposed, then you will end up with a kernel of the posterior density, but it may not integrate to one (and thus it is not generally a density).
